Alligation or Mixture Questions and Answers

Answer: D

Take the quantities as 1, 1 and 2 kg, four kilograms in all.

\[ 126 + 135 + 2p = 4 \times 153 = 612 \]

\[ 2p = 612 - 261 = 351 \quad\Longrightarrow\quad p = \text{Rs. } 175.50 \]

The third variety carries half the weight of the mixture, which is why its price does not have to be far above 153 to lift the average.

Answer: B

Let the jar initially contain 8 litres of mixture:3 litres water and 5 litres alcohol

Let x litres of this mixture is drawn out and is replaced by x litres of water.

Amount of water now in the solution: 3-(3/8)x+x.

Amount of alcohol now in the solution: 5-(5/8)x

Desired ratio: 1/1

=>3-(3/8)x+x=5-(5/8)x

x=8/5 litres

%x=((8/5)/8)*100=20%

Answer: C

Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.

Answer: C

The alligation rule gives the ratio in which two ingredients must be mixed:

\[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer} - \text{mean}}{\text{mean} - \text{cheaper}} \]

Treat the profit percentages as the two "prices".

\[ \frac{\text{8\% part}}{\text{18\% part}} = \frac{18 - 14}{14 - 8} = \frac{4}{6} = \frac{2}{3} \]

So the 1000 kg splits into 2 and 3 of 5 parts:

\[ \text{18\% part} = \frac{3}{5} \times 1000 = 600 \text{ kg} \]

The overall 14% is nearer 18 than 8, so the higher-profit portion must be the larger - a check on which share is which.

Answer: D

Correct answer: 56.89 litres

Each operation leaves the same fraction of wine behind, so after \( n \) operations the quantity remaining is \[ Q\left(1 - \frac{x}{Q}\right)^{n} = 81\left(1 - \frac{9}{81}\right)^{3} = 81 \times \left(\frac{8}{9}\right)^{3}. \] Computing, \[ 81 \times \frac{512}{729} = \frac{41472}{729} = 56.89 \text{ litres}. \] The reason a formula is needed rather than repeated subtraction is that after the first replacement the cask holds a mixture, so each later withdrawal removes water as well as wine. Subtracting 9 litres three times would give 54, which is the distractor and the commonest error. Following it through confirms the result: 72 litres of wine remain after the first replacement, 64 after the second, and \( 64 \times \frac{8}{9} = 56.89 \) after the third.

Answer: B

Let the jar initially contain 8 liters of mixture:3 liters water and 5 liters alcohol

Let x liters of this mixture is drawn out and is replaced by x liters of water.

Amount of water now in the solution: 3-(3/8)x+x.

Amount of alcohol now in the solution: 5-(5/8)x

Desired ratio: 1/1

=>3-(3/8)x+x=5-(5/8)x

x=8/5 liters

%x=((8/5)/8) × 100=20%

Answer: C

Water is free, so the total value of the bucket never changes; only the volume does.

\[ \text{value} = 40 \times 3.50 = \text{Rs. } 140 \]

At Rs. 2 per litre that value must spread over

\[ \frac{140}{2} = 70 \text{ litres} \]

\[ \text{water added} = 70 - 40 = 30 \text{ litres} \]

By alligation, water is priced at 0 and \( \tfrac{\text{milk}}{\text{water}} = \tfrac{2-0}{3.50-2} = \tfrac{4}{3} \), so 40 litres of milk needs 30 of water.

Answer: C

Correct answer: 120 litres

Let the capacity be \( V \) litres. The milk initially is \( \frac{3}{5}V \). Drawing off 20 litres of the mixture removes milk in the same proportion, that is \( \frac{3}{5} \times 20 = 12 \) litres, leaving \( \frac{3}{5}V - 12 \). Adding 20 litres of water returns the total to \( V \). For the final ratio to be 1 : 1 the milk must be half the total: \[ \frac{3}{5}V - 12 = \frac{V}{2} \;\Rightarrow\; \frac{3V}{5} - \frac{V}{2} = 12 \;\Rightarrow\; \frac{V}{10} = 12 \;\Rightarrow\; V = 120. \] Checking: 120 litres in the ratio 3 : 2 is 72 of milk and 48 of water. Drawing 20 litres removes 12 and 8, leaving 60 and 40; adding 20 litres of water gives 60 and 60, which is 1 : 1. The key step is that the portion drawn off carries both components in their current proportion.

Answer: A

Remove the profit first to get the cost price of the mixture.

\[ \text{CP} = \frac{68.20}{1.10} = \text{Rs. } 62 \]

\[ \frac{\text{Rs. 60 tea}}{\text{Rs. 65 tea}} = \frac{65 - 62}{62 - 60} = \frac{3}{2} \]

The mean must lie between the two prices; 68.20 does not, which is why it cannot be used directly.

Answer: D

The alligation rule gives the ratio in which two ingredients must be mixed:

\[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer} - \text{mean}}{\text{mean} - \text{cheaper}} \]

\[ \frac{\text{Rs. 8.70 sugar}}{\text{Rs. 9.70 sugar}} = \frac{9.70 - 9}{9 - 8.70} = \frac{0.70}{0.30} = \frac{7}{3} \]

The mean of 9 lies much nearer 8.70, so the cheaper sugar dominates - consistent with 7 parts to 3.